LeetCode 508. Most Frequent Subtree Sum (c++)

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

  5
 /  \
2   -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

  5
 /  \
2   -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

思路一：map + 一个额外的vector用来保存sum的顺序。

class Solution {
public:
    vector findFrequentTreeSum(TreeNode* root) {
        if (!root) return solution;
        dfs(root);

        for (vector::iterator it = order.begin(); it != order.end(); ++it) {
            int sum = *it;
            if(sum_map[sum] == max_num) {
                solution.push_back(sum);
                sum_map[sum] = 0;
            }
        }
 
        return solution;
    }
    
    int dfs(TreeNode* root) {
        if(!root) return 0;
        int sum = root->val;
        sum += dfs(root->left);
        sum += dfs(root->right);

        order.push_back(sum);
        ++sum_map[sum];
        if(sum_map[sum] > max_num) max_num = sum_map[sum];
        return sum;
    }
private:
    vector solution;
    vector order;
    map sum_map;
    int max_num = 0;
};

思路二：unordered_map

class Solution {
public:
    vector findFrequentTreeSum(TreeNode* root) {
        unordered_map counts;
        int maxCount = 0;
        countSubtreeSums(root, counts, maxCount);
        
        
        vector maxSums;
        for(const auto& x :  counts){
            if(x.second == maxCount) maxSums.push_back(x.first);
        }
        return maxSums;
    }
    
    int countSubtreeSums(TreeNode *r, unordered_map &counts, int& maxCount){
        if(r == nullptr) return 0;
        
        int sum = r->val;
        sum += countSubtreeSums(r->left, counts, maxCount);
        sum += countSubtreeSums(r->right, counts, maxCount);
        ++counts[sum];
        maxCount = max(maxCount, counts[sum]);
        return sum;
    }
};